All categories

3 years ago

Higher Order Thinking A yellow ribbon

Mathematics

1 answer:

Olegator [25]3 years ago

8 0

Answer:

112

Step-by-step explanation:

You might be interested in

Enter the indicated element.

Ymorist [56]

$a_{1,3} = 9$

Step-by-step explanation:

The elements of matrices are represented in the form

$a_{r,c}$

Here r is the row number and c is the column number.

Given matrix is:

$A = \left[\begin{array}{ccc}3&6&9\\2&4&8\end{array}\right]$

The elements of matrix will be represented by a_{r,c}

We have to find $a_{1,3}$

Which means that the first row and third column.

The first row is: 3 6 9 and the element in third column is: 9

So,

$a_{1,3} = 9$

Keywords: Matrices, Elements

Learn more about matrices at:

#LearnwithBrainly

6 0

3 years ago

Which value of n makes this equation true?

jonny [76]

Answer:

the answer is A. -16

Step-by-step explanation:

3(-16)+3= -45

-45/5=

-9

5(-16)-1=-81

-81/9=

-9

-9=-9

4 0

3 years ago

What is the vertex of the equation y = -x 2 + 10x - 28?

Roman55 [17]

Answer:

(5,-3)

Step-by-step explanation:

Rewrite the equation in vertex form

y =-(x-5)^2 - 3

You take the value of -5 (that would be your x value) and change it to the opposite sign -> +5

Then you take -3 (that would be your y value) and take it as it is

therefore the answer is (5, -3)

3 0

3 years ago

Evaluate the expression. 34⋅23

Alex17521 [72]

Answer:

784

Step-by-step explanation:

(30+4)(23+3)

7 0

4 years ago

i need help with this equation please there are two more possible answers that were cut off they are 17,2% and 19,5%

Margarita [4]

Consider that the experimental probability of an event is based upon the previous trials and observations of the experiment.

The experimental probability of occurrence of an event is given by,

$\text{Probability of an event}=\frac{\text{ Number of outcomes that favoured the event}}{\text{ Total number of trials or outcomes}}$

As per the problem, there are a total of 1230 trials of rolling a dice.

And the favourable event is getting a 2.

The corresponding experimental probability is calculated as,

$\begin{gathered} P(\text{ getting a 2})=\frac{\text{ No. of times 2 occurred}}{\text{ Total no. of times the dice is thrown}} \\ P(\text{ getting a 2})=\frac{172}{1230} \\ P(\text{ getting a 2})\approx0.13984 \\ P(\text{ getting a 2})\approx13.98\text{ percent} \end{gathered}$

Thus, the required probability is 13.98% approximately.

Theref

7 0

1 year ago